Solve (4x-3)^2=64

$$( 4 x - 3 ) ^ { 2 } = 64$$

Solve for x

$$x = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=64$$

Subtract 64 from both sides.

$$16x^{2}-24x+9-64=0$$

Subtract 64 from 9 to get -55.

$$16x^{2}-24x-55=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-55. To find a and b, set up a system to be solved.

$$a+b=-24 ab=16\left(-55\right)=-880$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -880.

$$1,-880 2,-440 4,-220 5,-176 8,-110 10,-88 11,-80 16,-55 20,-44 22,-40$$

Calculate the sum for each pair.

$$1-880=-879 2-440=-438 4-220=-216 5-176=-171 8-110=-102 10-88=-78 11-80=-69 16-55=-39 20-44=-24 22-40=-18$$

The solution is the pair that gives sum -24.

$$a=-44 b=20$$

Rewrite 16x^{2}-24x-55 as \left(16x^{2}-44x\right)+\left(20x-55\right).

$$\left(16x^{2}-44x\right)+\left(20x-55\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(4x-11\right)+5\left(4x-11\right)$$

Factor out common term 4x-11 by using distributive property.

$$\left(4x-11\right)\left(4x+5\right)$$

To find equation solutions, solve 4x-11=0 and 4x+5=0.

$$x=\frac{11}{4} x=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=64$$

Subtract 64 from both sides.

$$16x^{2}-24x+9-64=0$$

Subtract 64 from 9 to get -55.

$$16x^{2}-24x-55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -24 for b, and -55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 16\left(-55\right)}}{2\times 16}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 16\left(-55\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-24\right)±\sqrt{576-64\left(-55\right)}}{2\times 16}$$

Multiply -64 times -55.

$$x=\frac{-\left(-24\right)±\sqrt{576+3520}}{2\times 16}$$

Add 576 to 3520.

$$x=\frac{-\left(-24\right)±\sqrt{4096}}{2\times 16}$$

Take the square root of 4096.

$$x=\frac{-\left(-24\right)±64}{2\times 16}$$

The opposite of -24 is 24.

$$x=\frac{24±64}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{24±64}{32}$$

Now solve the equation x=\frac{24±64}{32} when ± is plus. Add 24 to 64.

$$x=\frac{88}{32}$$

Reduce the fraction \frac{88}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{11}{4}$$

Now solve the equation x=\frac{24±64}{32} when ± is minus. Subtract 64 from 24.

$$x=-\frac{40}{32}$$

Reduce the fraction \frac{-40}{32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{5}{4}$$

The equation is now solved.

$$x=\frac{11}{4} x=-\frac{5}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=64$$

Subtract 9 from both sides.

$$16x^{2}-24x=64-9$$

Subtract 9 from 64 to get 55.

$$16x^{2}-24x=55$$

Divide both sides by 16.

$$\frac{16x^{2}-24x}{16}=\frac{55}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{24}{16}\right)x=\frac{55}{16}$$

Reduce the fraction \frac{-24}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{3}{2}x=\frac{55}{16}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{55}{16}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{55+9}{16}$$

Add \frac{55}{16} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=4$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-\frac{3}{4}=2 x-\frac{3}{4}=-2$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{11}{4} x=-\frac{5}{4}$$